primitive of closed forms along a continuous curve Here is the theorem: Suppose $\omega$ is a closed differential form in open domain $\Omega$. There is $\gamma:[a,b]\to\Omega$, that defines a continuos curve. Then $\omega$ has a primitive along $\gamma$. The primitive is unique up tp a constant. We obtained this primitive along line by the fact that closed curve has local primitive for each point. 
I have trouble understanding what is the primitive exactly. 
For example $\frac{dz}{z}$ is a closed form in $\mathbb{C}/\{0\}$. For a circle $\gamma$ around $0$, there is a primitive of $\frac{dz}{z}$ along this circle by this theorem. So shouldn't $\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z}=0$? But this is impossible. So what is the primitive of $\frac{dz}{z}$ exactly? Is it not single valued? Then what does the "unique up to a constant" mean?
 A: It has a primitive along a closed curve $\gamma :[a, b] \to \Omega$. It might not has a primitive on a closed loop. For example, let $\omega = \frac {dz}{z}$ and $\gamma :[0, 2\pi] \to \mathbb C \setminus \{0\}$ be given by $\gamma(t) = e^{it}$. Thus when restricted to $\gamma$, 
$$\frac{dz}{z} = \frac{d(e^{it})}{e^{it}} = \frac{ie^{it} dt}{e^{it}} = idt $$
and $idt$ has a primitive on $\gamma$, which is $it + C$ for some constant $C$. Using this, we have 
$$\int_\gamma \frac{dz}{z} = \int_0^{2\pi} d(it+C) = i(2\pi +C) - iC = 2\pi i .$$
That is indeed nothing to do with complex analysis. In my opinion, it is better to think that the differential forms lives on the interval $[a,b]$ (or $\mathbb S^1$) after you restrict that to $\gamma$. 
All differential one form on $[a,b]$ is given by $\alpha  = f(t)dt$. There is always a primitive: define a function $g:[a,b] \to \mathbb R$ by
$$g(t) = \int_a^t f(s)ds + C,$$
then 
$$dg = g'(t)dt = f(t) dt = \alpha$$
and so $\alpha$ has a primitive. 
On the other hand, not all differential one form $\alpha$ on $\mathbb S^1$ has a primitive: All one form on $\mathbb S^1$ can be written as $\alpha(t) = f(t) dt$, where $f(t)$ is $2\pi $-periodic. For example, $\alpha= dt$ is a differential one form on $\mathbb S^1$ which has no primitive: Indeed, if there is a function $g:\mathbb S^1 \to \mathbb R$ so that $dg = dt$, then 
$$2\pi = \int_{\mathbb S^1} dt = \int_{\mathbb S^1 } dg = 0,$$
which is impossible. One can show that a one form $\alpha(t) = f(t) dt$ on $\mathbb S^1$ has a primitive if and only if 
$$\int_{\mathbb S^1} \alpha = \int_0^{2\pi} f(t) dt = 0.$$
